How many moles of ammonium nitrate are in 335 mL of 0.425M #NH_4NO_3#?

Answer 1

#"0.142 moles"#

Although it would not solve the problem, you could get fancy and claim that this solution contains zero moles of ammonium nitrate.

This is why it is.

The number of moles of solute per liter of solution is indicated by the molarity of the solution.

The problem with soluble ionic compounds is that they dissociate completely in aqueous solution to form cations and anions. In this case, ammonium nitrate exists in solution as ammonium cations, #"NH"_4^(+)#, and nitrate anions, #"NO"_3^(-)#
#"NH"_4"NO"_text(3(aq]) -> "NH"_text(4(aq])^(+) + "NO"_text(3(aq])^(-)#

As a result, the concentration of ammonium nitrate in your solution is theoretically zero since it doesn't exist.

Molarity, to put it simply, was first used to express the concentration of a specific chemical species that was present in solution. (For more on that, go here).

Nonetheless, molarity is now frequently used to express a solute's concentration in solutions, independent of the form the solute takes.

In your case, you know that the solution has a molariy of #"0.425 mol L"^(-1)# and a total volume of #"335 mL"#. To find how many moles of ammonium nitrate you have in solution, use the equation
#color(blue)(|bar(ul(color(white)(a/a)c = n_"solute"/V_"solution"color(white)(a/a)|)))#

Remember to use the conversion factor to convert the volume to liters.

#"1 L" = 10^3"mL"#

You'll possess

#c = n/V implies n = c * V#
#n = "0.425 mol" color(red)(cancel(color(black)("L"^(-1)))) * 335 * 10^(-3)color(red)(cancel(color(black)("L"))) = color(green)(|bar(ul(color(white)(a/a)"0.142 moles"color(white)(a/a)|)))#

Three sig figs are used to round the result.

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Answer 2

To calculate the number of moles of ammonium nitrate (NH4NO3) in 335 mL of a 0.425M solution, you use the formula:

moles = molarity × volume (in liters)

First, convert the volume from milliliters to liters:

335 mL ÷ 1000 = 0.335 liters

Now, use the formula:

moles = 0.425 mol/L × 0.335 L

moles ≈ 0.142 mol

So, there are approximately 0.142 moles of ammonium nitrate in 335 mL of a 0.425M NH4NO3 solution.

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Answer 3

To find the number of moles of ammonium nitrate in 335 mL of a 0.425 M solution, you would use the formula:

[ \text{moles} = \text{molarity} \times \text{volume} ]

Substituting the given values:

[ \text{moles} = 0.425 , \text{mol/L} \times 0.335 , \text{L} ]

[ \text{moles} = 0.142 , \text{mol} ]

So, there are 0.142 moles of ammonium nitrate in 335 mL of 0.425 M NH4NO3 solution.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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